A bag contains 5 green and 7 blue balls.another one contains 4 green and 8 blue balls.if a bag is choosen randomly and a ball is drawn from it at random,then find the probability that it is green:
(A)3/4 (B)1/4 (C)5/8 (D)3/8 (E)none of these

• (1/2*5/12)+(1/2*4/12)=3/8
• 13 years agoHelpfull: Yes(9) No(2)
• probability of choose a first bag=1/2,
  probability of draw a green ball from first bag=5/12,
  Now, probability of get green ball from first bag randomly=1/2 * 5/12=5/24.
  similarly, for second bag = 1/2 * 4/12=1/6.
  Therefore, total probability = 5/24 + 1/6 =3/8(d)
• 13 years agoHelpfull: Yes(5) No(0)
• sorry my mistake it would be 9/24=3/8
• 13 years agoHelpfull: Yes(3) No(0)
• AMITABH BARMAN check ur solution.....it will be 3/8.
• 13 years agoHelpfull: Yes(2) No(0)
• 1/2(5/12+4/12)=9/24=3/8
  ans=3/8
• 13 years agoHelpfull: Yes(2) No(0)
• probability that the ball is green will be first picking a ball form a bag whose probability itself is half and then picking up green ball from it and then same for the next bag
  probability of ball being green=1/2 x 5/12 + 1/2 x 4/12 =9/12=3/4 so then the answer is a
• 13 years agoHelpfull: Yes(1) No(7)
• D
• 13 years agoHelpfull: Yes(0) No(2)
• probability for a bag to be choosen = 1/2
  for first bag probability of green ball = 5/12
  for another bag probability of green ball = 4/12
  so , probability = (1/2*5/12)+(1/2*4/12) = 3/8
• 6 years agoHelpfull: Yes(0) No(0)